Difference between revisions of "User:Harold/AlgTop1"
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\ldots \ar[r] & H_k(S^{n-1}) \ar[r] \ar@2{->}[d]^{i_*} & H_k(B^n) \ar[r] \ar@2{->}[d]^{\text{Id}} & H_k(D^n, S^{n-1}) \ar[r] \ar@{.>}[d]^{i'_*}& H_{k-1}(S^{n-1}) \ar[r] \ar@2{->}[d]^{i_*} & H_{k-1}(B^n) \ar@2{->}[d]^{\text{Id} } \ar[r] & \ldots \\ | \ldots \ar[r] & H_k(S^{n-1}) \ar[r] \ar@2{->}[d]^{i_*} & H_k(B^n) \ar[r] \ar@2{->}[d]^{\text{Id}} & H_k(D^n, S^{n-1}) \ar[r] \ar@{.>}[d]^{i'_*}& H_{k-1}(S^{n-1}) \ar[r] \ar@2{->}[d]^{i_*} & H_{k-1}(B^n) \ar@2{->}[d]^{\text{Id} } \ar[r] & \ldots \\ | ||
\ldots \ar[r] & H_k(B^n-\{0\}) \ar[r] & H_k(B^n) \ar[r] & H_k(D^n, B^n-\{0\}) \ar[r] & H_{k-1}(B^n-\{0\}) \ar[r] & H_{k-1}(B^n) \ar[r] & ... | \ldots \ar[r] & H_k(B^n-\{0\}) \ar[r] & H_k(B^n) \ar[r] & H_k(D^n, B^n-\{0\}) \ar[r] & H_{k-1}(B^n-\{0\}) \ar[r] & H_{k-1}(B^n) \ar[r] & ... | ||
| + | }</m></span></center> | ||
| + | |- | ||
| + | ! Diagram | ||
| + | |} | ||
| + | <!-- | ||
| + | <harrynoob> ... -> H_k(B^n\0) -> H_k(B^n) -> H_k(B^n, B^n\0) -> H_k-1(B^n\0) -> H_k-1(B^n) -> ... | ||
| + | <harrynoob> ... -> H_k(R^n\0) -> H_k(R^n) -> H_k(R^n, R^n\0) -> H_k-1(R^n\0) -> H_k-1(R^n) -> ... | ||
| + | --> | ||
| + | {| class="wikitable" border="1" style="overflow:hidden;" | ||
| + | |- | ||
| + | | <center><span style="font-size:1.2em;"><m>\xymatrix{ | ||
| + | \ldots \ar[r] & H_k(B^n-\{0\}) \ar[r] \ar[d]^{\text{?} } & H_k(B^n) \ar[r] \ar[d]^{\text{?}} & H_k(B^n, B^n-\{0\}) \ar[r] \ar@{.>}[d]^{\text{?} }& H_{k-1}(B^n-\{0\}) \ar[r] \ar[d]^{\text{?} } & H_{k-1}(B^n) \ar[d]^{\text{?} } \ar[r] & \ldots \\ | ||
| + | \ldots \ar[r] & H_k(\mathbb{R}^n-\{0\}) \ar[r] & H_k(\mathbb{R}^n) \ar[r] & H_k(\mathbb{R}^n, R^n-\{0\}) \ar[r] & H_{k-1}(\mathbb{R}^n-\{0\}) \ar[r] & H_{k-1}(\mathbb{R}^n) \ar[r] & ... | ||
}</m></span></center> | }</m></span></center> | ||
|- | |- | ||
! Diagram | ! Diagram | ||
|} | |} | ||
Latest revision as of 22:57, 21 February 2017
Problem statement
Let n≥1. We show that the map Hk(Bn,Sn−1)→Hk(Bn,Bn∖{0}) by the inclusion i:Sn−1→Bn∖{0} is an isomorphism.
Tools
First we show that the map Hk(Sn−1)→Hk(Bn∖{0}) induced by the inclusion i:Sn−1→Bn∖{0} is an isomorphism, for k≥0. Note that Sn−1 is a retract of Bn∖{0}.
- Define the map r:Bn∖{0}→Sn−1 by r:x↦x||x||, where ||x|| denotes the norm of x.
- Then r∘i=idSn−1.
- So i∗:Sn−1→Bn∖{0} is a monomorphism.
- Also, i∘r is homotopy equivalent to the identity map on Bn∖{0}. [left as an exercise to the reader]
